Answer :
To find the common difference and the first term of the arithmetic sequence given that the second term is 24 and the fifth term is 3, we will use the formula for the [tex]\( n \)[/tex]-th term of an arithmetic sequence, which is:
[tex]\[ a_n = a_1 + (n-1)d \][/tex]
where:
- [tex]\( a_n \)[/tex] is the [tex]\( n \)[/tex]-th term,
- [tex]\( a_1 \)[/tex] is the first term,
- [tex]\( d \)[/tex] is the common difference,
- [tex]\( n \)[/tex] is the term number.
Given:
- [tex]\( a_2 = 24 \)[/tex]
- [tex]\( a_5 = 3 \)[/tex]
Using the formula for these terms, we can set up the following equations:
For the second term ([tex]\( n = 2 \)[/tex]):
[tex]\[ a_2 = a_1 + (2-1)d \][/tex]
[tex]\[ 24 = a_1 + d \][/tex] [tex]\[ \quad (1) \][/tex]
For the fifth term ([tex]\( n = 5 \)[/tex]):
[tex]\[ a_5 = a_1 + (5-1)d \][/tex]
[tex]\[ 3 = a_1 + 4d \][/tex] [tex]\[ \quad (2) \][/tex]
We now have a system of linear equations:
[tex]\[ 24 = a_1 + d \quad \text{(1)} \][/tex]
[tex]\[ 3 = a_1 + 4d \quad \text{(2)} \][/tex]
To eliminate [tex]\( a_1 \)[/tex], we can subtract equation (1) from equation (2):
[tex]\[ (a_1 + 4d) - (a_1 + d) = 3 - 24 \][/tex]
[tex]\[ a_1 + 4d - a_1 - d = 3 - 24 \][/tex]
[tex]\[ 3d = -21 \][/tex]
[tex]\[ d = -7 \][/tex]
The common difference [tex]\( d \)[/tex] is [tex]\(-7\)[/tex].
Next, we substitute [tex]\( d = -7 \)[/tex] back into equation (1) to find the first term [tex]\( a_1 \)[/tex]:
[tex]\[ 24 = a_1 + (-7) \][/tex]
[tex]\[ 24 = a_1 - 7 \][/tex]
[tex]\[ a_1 = 24 + 7 \][/tex]
[tex]\[ a_1 = 31 \][/tex]
Therefore, the common difference [tex]\( d \)[/tex] is [tex]\(-7\)[/tex], and the first term [tex]\( a_1 \)[/tex] is [tex]\( 31 \)[/tex].
[tex]\[ a_n = a_1 + (n-1)d \][/tex]
where:
- [tex]\( a_n \)[/tex] is the [tex]\( n \)[/tex]-th term,
- [tex]\( a_1 \)[/tex] is the first term,
- [tex]\( d \)[/tex] is the common difference,
- [tex]\( n \)[/tex] is the term number.
Given:
- [tex]\( a_2 = 24 \)[/tex]
- [tex]\( a_5 = 3 \)[/tex]
Using the formula for these terms, we can set up the following equations:
For the second term ([tex]\( n = 2 \)[/tex]):
[tex]\[ a_2 = a_1 + (2-1)d \][/tex]
[tex]\[ 24 = a_1 + d \][/tex] [tex]\[ \quad (1) \][/tex]
For the fifth term ([tex]\( n = 5 \)[/tex]):
[tex]\[ a_5 = a_1 + (5-1)d \][/tex]
[tex]\[ 3 = a_1 + 4d \][/tex] [tex]\[ \quad (2) \][/tex]
We now have a system of linear equations:
[tex]\[ 24 = a_1 + d \quad \text{(1)} \][/tex]
[tex]\[ 3 = a_1 + 4d \quad \text{(2)} \][/tex]
To eliminate [tex]\( a_1 \)[/tex], we can subtract equation (1) from equation (2):
[tex]\[ (a_1 + 4d) - (a_1 + d) = 3 - 24 \][/tex]
[tex]\[ a_1 + 4d - a_1 - d = 3 - 24 \][/tex]
[tex]\[ 3d = -21 \][/tex]
[tex]\[ d = -7 \][/tex]
The common difference [tex]\( d \)[/tex] is [tex]\(-7\)[/tex].
Next, we substitute [tex]\( d = -7 \)[/tex] back into equation (1) to find the first term [tex]\( a_1 \)[/tex]:
[tex]\[ 24 = a_1 + (-7) \][/tex]
[tex]\[ 24 = a_1 - 7 \][/tex]
[tex]\[ a_1 = 24 + 7 \][/tex]
[tex]\[ a_1 = 31 \][/tex]
Therefore, the common difference [tex]\( d \)[/tex] is [tex]\(-7\)[/tex], and the first term [tex]\( a_1 \)[/tex] is [tex]\( 31 \)[/tex].